Tag: unsolved

It is a well-known open problem whether the following process terminates for every positive integer:

3n+1 flow chart

Experiments suggest that it does, possibly with unintended side effects.

Since for any odd integer the number is even, it is convenient to replace the step with , as show below:

(3n+1)/2 flow chart

As a function of , the stopping time has a nice patterned graph:

Stopping time

An odd integer is of the form or . In the first case, is even, while in the second case is odd. So, if is picked randomly from all odd integers of certain size, the probability of being even is . Similarly, for an even , the number can be even or odd with probability .

This leads to a stochastic model of the process:

Stochastic flow

The graph of stopping time in the stochastic model is, of course random. It looks nothing like the nice pattern of the deterministic process.

Stopping time, stochastic version

However, smoothing out both graphs by moving window of width or so, we see the similarity:

Moving averages, deterministic and stochastic

The stochastic process is much easier to analyze. Focusing on the logarithm , we see that it changes either by or by approximately . The expected value of the change is . This suggests that we can expect the logarithm to drop down to in about steps. (Rigorous derivation requires more tools from probability theory, but is still routine.)

The curve fits the experimental data nicely. (The red curve, being randomly generated, is different from the one on the previous graph.)

For the computations, I used Scilab. The function hail(n,m) calculates the stopping times up to given value of n, and takes moving average with window size m (which can be set to 1 for no averaging).

function hail(n,m)
steps=zeros(1:n);
steps(1)=0
for i=2:n
k=i;
s=0;
while k>=i
if modulo(k,2)==0 then
k=k/2;
s=s+1;
else
k=(3*k+1)/2;
s=s+1;
end
end
steps(i)=s+steps(k);
end
total = cumsum(steps)
for i=1:n-m
average(i)=(total(i+m)-total(i))/m;
end
plot(average,'+');
endfunction

As soon as the result of computations drops below the starting value, the number of remaining steps is fetched from the array that is already computed. This speeds up the process a bit.

The second function follows the stochastic model, for which the aforementioned optimization is not available. This is actually an interesting point: it is conceivable that the stochastic model would be more accurate if it also used the pre-computed stopping time once drops below the starting value. This would change the distribution of stopping times, resulting in wider fluctuations after averaging.

Good news first. Given a 2-dimensional subspace of (or ) such that (), we can always find a coordinate subspace such that and are in generic position. (Recall that generic position means , which is feasible only when the dimension of and is half the dimension of the ambient space.)

Indeed, suppose that none of the subspaces , , are in generic position with . By the pigeonhole principle, either or meets at least two of these subspaces. We may assume meets and . Since is two-dimensional, it follows that . Hence , a contradiction.

The story is different in and higher dimensions. Consider the space

, with

The condition holds. Yet, for any 3-dimensional coordinate space , either or its complement contains at least two of vectors , and therefore intersect . Damn perfidious pigeons.

So it’s not enough to demand that does not contain any basis vectors. Let’s ask it to stay away from the basis vectors, as far as possible. By the Pythagorean theorem, the maximal possible distance of to is , attained when is equidistant from and . Let’s call an equidistant subspace if this holds for all . There are at least two other natural ways to express this property:

In the basis , the projection onto is a matrix with 1/2 on the diagonal

Reflection across sends into a vector orthogonal to . As a matrix, this reflection has 0s on the diagonal.

In two dimensions, the only equidistant subspaces are and . In higher dimensions they form a connected subset of the Grassmannian (self-promotion).

Is every equidistant subspace in generic position with some coordinate subspace?

We already saw that the answer is yes when . It is also affirmative when (G. Weiss and V. Zarikian, “Paving small matrices and the Kadison-Singer extension problem”, 2010). I am 75% sure that the answer is yes in general.

The sequences in The On-Line Encyclopedia of Integer Sequences® are themselves arranged into a sequence, each being indexed by Axxxxxx. Hence, the sequence2, 3, 2, 1, 3, 4, 1, 7, 7, 5, 45, 2, 181, 43, 17, ...
can never be included in the encyclopedia: its definition is .

By the way, which sequences have the honor of being on the virtual first page of the OEIS? A000004 consists of zeros. Its description offers the following explicit formula:

This is the Kolakoski sequence, which first appeared in print in 1965. Contrary to what one might expect from such a simple definition, there is no simple pattern; it is not even known whether 1s and 2s appear with the same asymptotic frequency 1/2. One way to look for patterns in a function/sequence is to take its Fourier transform, and this is what I did below:

Fourier transform: click to magnify

Specifically, this is the absolute value of . To better see where the peaks fall, it’s convenient to look at :

Reciprocal axis: click to magnify

Something is going on at , i.e. at period 3. Recall that the sequence contains no triples 111 or 222: there are no runs of more than 2 equal numbers. So perhaps it’s not surprising that a pattern emerges at period 3. And indeed, even though the percentage of 1s among the first 500 terms of the sequence is 49.8%, breaking it down by n mod 3 yields the following:

Among , the percentage of 1s is 17.4%

Among , the percentage of 1s is 85.2%

Among , the percentage of 1s is 46.8%

Noticing another peak at , I also broke down the percentages by n mod 9:

: there are 5.4% 1s

: there are 74.8% 1s

: there are 41.4% 1s

: there are 21.6% 1s

: there are 97.2% 1s

: there are 70.2% 1s

: there are 25.2% 1s

: there are 84.6% 1s

: there are 28.8% 1s

Nothing close to 50%…

In conclusion, my code for the Kolakoski sequence. This time I used Maple (instead of Scilab) for no particular reason.

Suppose is a polynomial which we can evaluate at the th roots of unity. By way of normalization, assume that whenever . What is the best upper estimate for on the unit circle, i.e., the smallest number for which we can say ?

There is no bound at all if , because can vanish at all roots of unity and be as large elsewhere as it pleases. So the first nontrivial case is . In this case our is the Lagrange interpolation polynomial with equidistributed nodes on the unit circle. Since its values at the nodes are bounded by 1, the matter reduces to estimating the basis polynomials where are the nodes. This leads to geometrically appealing calculations such as

N black dots uniformly distributed on the unit circle

and

N black dots uniformly distributed on the unit circle; and a green midpoint

After summation we get a partial sum of the harmonic series, and so is of order . I’ve seen this result attributed to Marcinkiewicz, specifically to his 1937 paper Sur la divergence de polynomes d’interpolation. However, I could not find anything of the sort in this paper. The estimate is used, without proof, in his other paper in the same issue of the journal: Quelques remarques sur l’interpolation. I’m pretty sure Marcinkiewicz did not consider the estimate new; most likely, Bernstein and Faber did such computations earlier.

Anyway, what happens if ? Fast forward to the 21 century. In the 2005 paper On discrete norms of polynomials by Рахманов and Шехтман, the constant is shown to be of order for all . The logarithm matters only if is close to 1; for large values (say, ) this estimate is simply with implicit constants . Keeping fixed, we should have as , but the estimate does not tell us this.

Shortly thereafter (2008), T. Sheil-Small obtained a clean estimate , which indeed has the property as . Interestingly, the estimate is sharp when (and only in this case): a multiple of achieves the value . The extremal polynomial has a zero at the midpoint of every other gap between the th roots of unity.

And quite recently (2011), В.Н. Дубинин obtained the estimate which is sharp whenever divides (and only then). In particular, that was secretly . The polynomial that achieves the value has a zero at the midpoint of every th gap between the th roots of unity.

Can one compute when is not divisible by ? I don’t want to say it’s impossible, but one would first need to understand the extremal polynomials, and my numerical experiments show no clear pattern. Their zeros do not lie on the unit circle; they are sometimes inside and sometimes outside. So I’m not optimistic.

Let’s say that a set contains an singular matrix if there exists an matrix with determinant zero whose entries are distinct elements of . For example, the set of prime numbers does not contain any singular matrices; however, for every it contains infinitely many singular matrices.

I don’t know of an elementary proof of the latter fact. By a 1939 theorem of van der Corput, the set of primes contains infinitely many progressions of length 3. (Much more recently, Green and Tao replaced 3 with an arbitrary .) If every row of a matrix begins with a 3-term arithmetic progression, the matrix is singular.

Let be a real Banach space with the dual . Fix and call a linear functional almost norming for if and . In Part 1 I showed that in any Banach space there exists a continuous selection of almost norming functionals. Here I will prove that there is no uniformly continuous selection in .

Claim. Let be the unit sphere in , the -dimensional -space. Suppose that is a map such that is almost norming in the above sense. Then the modulus of continuity satisfies .

(If an uniformly continuous selection was available in , it would yield selections in with a modulus of continuity independent of .)

Proof. Write . For any we have , hence

for all . Sum over all and change the order of summation:

There exists such that

Fix this from now on. Define to be the same vector as , but with the th component flipped. Rewrite the previous sum as

A map is -bilipschitz if for all . This definition makes sense if X and Y are general metric spaces, but let’s suppose they are subsets on the plane .

Definition 1. A set has the BL extension property if any bilipschitz map can be extended to a bilipschitz map . (Extension means that is required to agree with on .)

Lines and circles have the BL extension property. This was proved in early 1980s independently by Tukia, Jerison and Kenig, and Latfullin.

Definition 2. A set has the controlled BL extension property if there exists a constant such that any -bilipschitz map can be extended to a -bilipschitz map .

Clearly, Definition 2 asks for more than Definition 1. I can prove that a line has the controlled BL extension property, even with a modest constant such as . (Incidentally, one cannot take .) I still can’t prove the controlled BL extension property for a circle.